LeetCode #2512 — MEDIUM

Reward Top K Students

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given two string arrays positive_feedback and negative_feedback, containing the words denoting positive and negative feedback, respectively. Note that no word is both positive and negative.

Initially every student has 0 points. Each positive word in a feedback report increases the points of a student by 3, whereas each negative word decreases the points by 1.

You are given n feedback reports, represented by a 0-indexed string array report and a 0-indexed integer array student_id, where student_id[i] represents the ID of the student who has received the feedback report report[i]. The ID of each student is unique.

Given an integer k, return the top k students after ranking them in non-increasing order by their points. In case more than one student has the same points, the one with the lower ID ranks higher.

Example 1:

Input: positive_feedback = ["smart","brilliant","studious"], negative_feedback = ["not"], report = ["this student is studious","the student is smart"], student_id = [1,2], k = 2
Output: [1,2]
Explanation: 
Both the students have 1 positive feedback and 3 points but since student 1 has a lower ID he ranks higher.

Example 2:

Input: positive_feedback = ["smart","brilliant","studious"], negative_feedback = ["not"], report = ["this student is not studious","the student is smart"], student_id = [1,2], k = 2
Output: [2,1]
Explanation: 
- The student with ID 1 has 1 positive feedback and 1 negative feedback, so he has 3-1=2 points. 
- The student with ID 2 has 1 positive feedback, so he has 3 points. 
Since student 2 has more points, [2,1] is returned.

Constraints:

1 <= positive_feedback.length, negative_feedback.length <= 10⁴
1 <= positive_feedback[i].length, negative_feedback[j].length <= 100
Both positive_feedback[i] and negative_feedback[j] consists of lowercase English letters.
No word is present in both positive_feedback and negative_feedback.
n == report.length == student_id.length
1 <= n <= 10⁴
report[i] consists of lowercase English letters and spaces ' '.
There is a single space between consecutive words of report[i].
1 <= report[i].length <= 100
1 <= student_id[i] <= 10⁹
All the values of student_id[i] are unique.
1 <= k <= n

Step 01

Brute Force Baseline

Problem summary: You are given two string arrays positive_feedback and negative_feedback, containing the words denoting positive and negative feedback, respectively. Note that no word is both positive and negative. Initially every student has 0 points. Each positive word in a feedback report increases the points of a student by 3, whereas each negative word decreases the points by 1. You are given n feedback reports, represented by a 0-indexed string array report and a 0-indexed integer array student_id, where student_id[i] represents the ID of the student who has received the feedback report report[i]. The ID of each student is unique. Given an integer k, return the top k students after ranking them in non-increasing order by their points. In case more than one student has the same points, the one with the lower ID ranks higher.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

["smart","brilliant","studious"]
["not"]
["this student is studious","the student is smart"]
[1,2]
2

Example 2

["smart","brilliant","studious"]
["not"]
["this student is not studious","the student is smart"]
[1,2]
2

Core Insight

What unlocks the optimal approach

Hash the positive and negative feedback words separately.
Calculate the points for each student’s feedback.
Sort the students accordingly to find the top <em>k</em> among them.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2512: Reward Top K Students
class Solution {
    public List<Integer> topStudents(String[] positive_feedback, String[] negative_feedback,
        String[] report, int[] student_id, int k) {
        Set<String> ps = new HashSet<>();
        Set<String> ns = new HashSet<>();
        for (var s : positive_feedback) {
            ps.add(s);
        }
        for (var s : negative_feedback) {
            ns.add(s);
        }
        int n = report.length;
        int[][] arr = new int[n][0];
        for (int i = 0; i < n; ++i) {
            int sid = student_id[i];
            int t = 0;
            for (var s : report[i].split(" ")) {
                if (ps.contains(s)) {
                    t += 3;
                } else if (ns.contains(s)) {
                    t -= 1;
                }
            }
            arr[i] = new int[] {t, sid};
        }
        Arrays.sort(arr, (a, b) -> a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]);
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < k; ++i) {
            ans.add(arr[i][1]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2512: Reward Top K Students
func topStudents(positive_feedback []string, negative_feedback []string, report []string, student_id []int, k int) (ans []int) {
	ps := map[string]bool{}
	ns := map[string]bool{}
	for _, s := range positive_feedback {
		ps[s] = true
	}
	for _, s := range negative_feedback {
		ns[s] = true
	}
	arr := [][2]int{}
	for i, sid := range student_id {
		t := 0
		for _, w := range strings.Split(report[i], " ") {
			if ps[w] {
				t += 3
			} else if ns[w] {
				t -= 1
			}
		}
		arr = append(arr, [2]int{t, sid})
	}
	sort.Slice(arr, func(i, j int) bool { return arr[i][0] > arr[j][0] || (arr[i][0] == arr[j][0] && arr[i][1] < arr[j][1]) })
	for _, v := range arr[:k] {
		ans = append(ans, v[1])
	}
	return
}

# Accepted solution for LeetCode #2512: Reward Top K Students
class Solution:
    def topStudents(
        self,
        positive_feedback: List[str],
        negative_feedback: List[str],
        report: List[str],
        student_id: List[int],
        k: int,
    ) -> List[int]:
        ps = set(positive_feedback)
        ns = set(negative_feedback)
        arr = []
        for sid, r in zip(student_id, report):
            t = 0
            for w in r.split():
                if w in ps:
                    t += 3
                elif w in ns:
                    t -= 1
            arr.append((t, sid))
        arr.sort(key=lambda x: (-x[0], x[1]))
        return [v[1] for v in arr[:k]]

// Accepted solution for LeetCode #2512: Reward Top K Students
use std::collections::{HashMap, HashSet};
impl Solution {
    pub fn top_students(
        positive_feedback: Vec<String>,
        negative_feedback: Vec<String>,
        report: Vec<String>,
        student_id: Vec<i32>,
        k: i32,
    ) -> Vec<i32> {
        let n = student_id.len();
        let ps = positive_feedback.iter().collect::<HashSet<&String>>();
        let ns = negative_feedback.iter().collect::<HashSet<&String>>();
        let mut map = HashMap::new();
        for i in 0..n {
            let id = student_id[i];
            let mut count = 0;
            for s in report[i].split(' ') {
                let s = &s.to_string();
                if ps.contains(s) {
                    count += 3;
                } else if ns.contains(s) {
                    count -= 1;
                }
            }
            map.insert(id, count);
        }
        let mut t = map.into_iter().collect::<Vec<(i32, i32)>>();
        t.sort_by(|a, b| {
            if a.1 == b.1 {
                return a.0.cmp(&b.0);
            }
            b.1.cmp(&a.1)
        });
        t.iter().map(|v| v.0).collect::<Vec<i32>>()[0..k as usize].to_vec()
    }
}

// Accepted solution for LeetCode #2512: Reward Top K Students
function topStudents(
    positive_feedback: string[],
    negative_feedback: string[],
    report: string[],
    student_id: number[],
    k: number,
): number[] {
    const n = student_id.length;
    const map = new Map<number, number>();
    const ps = new Set(positive_feedback);
    const ns = new Set(negative_feedback);
    for (let i = 0; i < n; i++) {
        map.set(
            student_id[i],
            report[i].split(' ').reduce((r, s) => {
                if (ps.has(s)) {
                    return r + 3;
                }
                if (ns.has(s)) {
                    return r - 1;
                }
                return r;
            }, 0),
        );
    }
    return [...map.entries()]
        .sort((a, b) => {
            if (a[1] === b[1]) {
                return a[0] - b[0];
            }
            return b[1] - a[1];
        })
        .map(v => v[0])
        .slice(0, k);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n + (|ps| + |ns| + n)

Space

O((|ps|+|ns|)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.